negative b over 2 plus or minus the square root of, uh, something
December 6, 2019 8:14 PM Subscribe
How to solve quadratic equations with a method (slmb*) instead of just using the quadratic formula. This helps folks like me who more easily remember steps than formulas.
*single link math blog
It's a shame that students are taught to memorize the quadratic formula, but are not taught the far more general method of completing the square which the formula is derived from. (Loh's method is essentially equivalent to completing the square.)
posted by a car full of lions at 8:56 PM on December 6, 2019 [5 favorites]
posted by a car full of lions at 8:56 PM on December 6, 2019 [5 favorites]
the far more general method of completing the square which the formula is derived from
When I was taught the formula somewhere back in the mists of time, it was presented as the endpoint of a bunch of completing-the-square approaches to various special cases of quadratic equations, with the specialness of the cases decreasing at each step and the formula getting correspondingly more complicated.
I remember being super impressed with the idea that a single formula could capture and bypass all this busywork and avoid any need to identify which special case was applicable, and the fact that it contained a subpart that could tell you whether any solution existed without needing to work out the rest. It was that sense of the power of that formula that wedged it firmly into my brain, where it remains to this day despite the fact that I've only ever had about two occasions to use it. Minus b plus or minus the square root of b squared minus four ac all over two a! Bam. Job done.
I had a good third form maths teacher.
posted by flabdablet at 9:31 PM on December 6, 2019 [14 favorites]
When I was taught the formula somewhere back in the mists of time, it was presented as the endpoint of a bunch of completing-the-square approaches to various special cases of quadratic equations, with the specialness of the cases decreasing at each step and the formula getting correspondingly more complicated.
I remember being super impressed with the idea that a single formula could capture and bypass all this busywork and avoid any need to identify which special case was applicable, and the fact that it contained a subpart that could tell you whether any solution existed without needing to work out the rest. It was that sense of the power of that formula that wedged it firmly into my brain, where it remains to this day despite the fact that I've only ever had about two occasions to use it. Minus b plus or minus the square root of b squared minus four ac all over two a! Bam. Job done.
I had a good third form maths teacher.
posted by flabdablet at 9:31 PM on December 6, 2019 [14 favorites]
Completing the square? You mean common core standard HSA REI B4a? One of the standards that is being taught to all high school students at schools that implement common core?
posted by mwalimu at 9:31 PM on December 6, 2019 [15 favorites]
posted by mwalimu at 9:31 PM on December 6, 2019 [15 favorites]
I see Loh also discusses completing the square in his more detailed article posted on arXiv.
posted by a car full of lions at 9:56 PM on December 6, 2019 [2 favorites]
posted by a car full of lions at 9:56 PM on December 6, 2019 [2 favorites]
Yes, I'm pretty sure completing the square is routinely taught in the US and has been for a while.
posted by hoyland at 10:02 PM on December 6, 2019 [2 favorites]
posted by hoyland at 10:02 PM on December 6, 2019 [2 favorites]
Hm... I don't remember learning about completing the square in my algebra class. I suspect that they spent the minimum amount of time possible on it before deriving the quadratic formula and going from there.
I *do* remember guess-and-check though. Mostly because I thought it was a BS way to do math, and felt completely justified in my opinion once we learned the quadratic formula.
posted by Aleyn at 10:22 PM on December 6, 2019 [1 favorite]
I *do* remember guess-and-check though. Mostly because I thought it was a BS way to do math, and felt completely justified in my opinion once we learned the quadratic formula.
posted by Aleyn at 10:22 PM on December 6, 2019 [1 favorite]
the negative boy couldn't decide to go a radical party the so he chose to be square and missed out on 4 awesome chicks and the party was over by 2am.
posted by lkc at 10:25 PM on December 6, 2019 [9 favorites]
posted by lkc at 10:25 PM on December 6, 2019 [9 favorites]
That average substitution idea is really neat. My math is rusty, so I blatantly cheated using calculus to find the optimum value, y*, of a quadratic function, and note that the two zeros lie ±w away from its symmetry axis such that aw2 = |y*|. The quadratic formula pops out.
But then I came to something weird that I don't get. The article example is x2–2x–24 = 0. If you argue that there exists some equivalent s, t such that (x–s)(x–t) = 0, you get the sum-and-product equations s + t = 2, st = –24, which then requires author's z substitution. However, if instead you argue that there exists a unique quadratic representation f(x) = (x–p)2+q, then its expansion and term-by-term matching gives the different system {p = 1, p2+q = 24} and the solvable constraint x = p ± √ q quickly gets you x = 6 or –4. Why is that?
posted by polymodus at 1:15 AM on December 7, 2019 [1 favorite]
But then I came to something weird that I don't get. The article example is x2–2x–24 = 0. If you argue that there exists some equivalent s, t such that (x–s)(x–t) = 0, you get the sum-and-product equations s + t = 2, st = –24, which then requires author's z substitution. However, if instead you argue that there exists a unique quadratic representation f(x) = (x–p)2+q, then its expansion and term-by-term matching gives the different system {p = 1, p2+q = 24} and the solvable constraint x = p ± √ q quickly gets you x = 6 or –4. Why is that?
posted by polymodus at 1:15 AM on December 7, 2019 [1 favorite]
The last bit should say … = –24, and ± √ –q.
posted by polymodus at 1:28 AM on December 7, 2019 [1 favorite]
posted by polymodus at 1:28 AM on December 7, 2019 [1 favorite]
500 years of not teaching the cubic formula.
posted by StephenB at 2:09 AM on December 7, 2019 [5 favorites]
posted by StephenB at 2:09 AM on December 7, 2019 [5 favorites]
> Why is that?
This is a hard question to answer, because why do you think it should be anything different?
These are two different methods and even though they are equivalent on some level, the exact steps are different and they show different values at different points because they are different.
More specifically, the crux of the method isn't finding systems of two equations and two unknowns and then solving them. The crux is finding a way to get that pesky square root handled.
So this pair of equations is true but not useful:
s + t = 2, st = -24
So two equations, two unknowns. Let's solve this. Rearrange the first equation to t=2-s and substitute that into the second equation:
s(2-s) = - 24
2s-s2=-24
Rearrange:
s2 - 2s - 24 = 0
So that's very true but not useful at all - it's the same thing we started with.
The "z substitution" is the trick that allows us to crack the square root problem, which is the nub of the whole method. It's the equivalent insight to the "complete the square" trick in the compete the square method.*
But it's coming at it from a slightly different direction and it's quite clever in its own way.
(*The insight of the Complete the Square method is that you can arrange the elements of the equation in such a way that they are like a square with one corner missing. We know the length of the side of the square and the size of the other pieces, so it's easy to figure out the exact size of the missing corner piece. Add that corner piece in and suddenly you can take the square root and instantly solve the problem. See the linked page for diagrams that make the whole thing quite clear.)
posted by flug at 2:53 AM on December 7, 2019 [2 favorites]
This is a hard question to answer, because why do you think it should be anything different?
These are two different methods and even though they are equivalent on some level, the exact steps are different and they show different values at different points because they are different.
More specifically, the crux of the method isn't finding systems of two equations and two unknowns and then solving them. The crux is finding a way to get that pesky square root handled.
So this pair of equations is true but not useful:
s + t = 2, st = -24
So two equations, two unknowns. Let's solve this. Rearrange the first equation to t=2-s and substitute that into the second equation:
s(2-s) = - 24
2s-s2=-24
Rearrange:
s2 - 2s - 24 = 0
So that's very true but not useful at all - it's the same thing we started with.
The "z substitution" is the trick that allows us to crack the square root problem, which is the nub of the whole method. It's the equivalent insight to the "complete the square" trick in the compete the square method.*
But it's coming at it from a slightly different direction and it's quite clever in its own way.
(*The insight of the Complete the Square method is that you can arrange the elements of the equation in such a way that they are like a square with one corner missing. We know the length of the side of the square and the size of the other pieces, so it's easy to figure out the exact size of the missing corner piece. Add that corner piece in and suddenly you can take the square root and instantly solve the problem. See the linked page for diagrams that make the whole thing quite clear.)
posted by flug at 2:53 AM on December 7, 2019 [2 favorites]
This is a hard question to answer, because why do you think it should be anything different?
I guess what piques my interest is that by assuming different representations, the solutions can be derived but using different numbers of formal steps (so, computational efficiency). Both (x–s)(x–t) and (x–p)2+q are general forms for the quadratic polynomial with coefficients a, b, c, but I don't know/remember their proofs.
Actually, this clarifies a different question: the author's approach does not seem to lend itself to a formal elementary proof, because a key step in it requires assuming that there exists a product of binomials that is equivalent to the quadratic polynomial. In contrast, completing the square is purely axiomatic: it makes no such assumptions but simply adds 0 as an expression, and thus gives a direct formal proof of the solutions.
posted by polymodus at 4:55 AM on December 7, 2019
I guess what piques my interest is that by assuming different representations, the solutions can be derived but using different numbers of formal steps (so, computational efficiency). Both (x–s)(x–t) and (x–p)2+q are general forms for the quadratic polynomial with coefficients a, b, c, but I don't know/remember their proofs.
Actually, this clarifies a different question: the author's approach does not seem to lend itself to a formal elementary proof, because a key step in it requires assuming that there exists a product of binomials that is equivalent to the quadratic polynomial. In contrast, completing the square is purely axiomatic: it makes no such assumptions but simply adds 0 as an expression, and thus gives a direct formal proof of the solutions.
posted by polymodus at 4:55 AM on December 7, 2019
Thanks! As a guy who dabbles in signals and systems, this will come in handy.
posted by ZenMasterThis at 6:07 AM on December 7, 2019 [1 favorite]
posted by ZenMasterThis at 6:07 AM on December 7, 2019 [1 favorite]
"Completing the square" is such a satisfying phrase. Surpassed, though, by "casting out nines", which makes it sound like you are doing witchcraft instead of arithmetic.
posted by thelonius at 6:50 AM on December 7, 2019 [5 favorites]
posted by thelonius at 6:50 AM on December 7, 2019 [5 favorites]
Yeah, I definitely don't do formal proofs much anymore but isn't Loh mistaken when they say this method circumvents and/or helps prove the fundamental theorem of algebra? I thought that asserting that the quadratic is factorable into binomials implicitly assumes the fundamental theorem but maybe I'm misunderstanding. It's a cool insight but I have to say the writing didn't help. I had to keep going back to the equations to understand what the sentences were doing which is usually not ideal. A lot of the time it seems like Loh is having a meta-argument with imagined naysayers rather than just presenting the method, which definitely set off my Crank Radar even though the results are good.
posted by range at 7:15 AM on December 7, 2019 [2 favorites]
posted by range at 7:15 AM on December 7, 2019 [2 favorites]
P.S. To the folks snarking at me about common core, I apologize for not being American.
posted by a car full of lions at 8:15 AM on December 7, 2019 [5 favorites]
posted by a car full of lions at 8:15 AM on December 7, 2019 [5 favorites]
Completing the square is taught in part so that you can find the center of a circle given its equation as
X^2 +AX + Y^2 + BY + C = 0
posted by Obscure Reference at 8:20 AM on December 7, 2019 [1 favorite]
X^2 +AX + Y^2 + BY + C = 0
posted by Obscure Reference at 8:20 AM on December 7, 2019 [1 favorite]
I was taught completing the square in Australia, and my kids are taught it today. I don't think they were explicitly taught that the b- coefficient is twice the average value of the roots; that seems a useful way to add clarity.
posted by Joe in Australia at 2:02 PM on December 7, 2019 [1 favorite]
posted by Joe in Australia at 2:02 PM on December 7, 2019 [1 favorite]
Metafilter: having a meta-argument with imagined naysayers .
posted by zengargoyle at 3:16 PM on December 7, 2019 [1 favorite]
posted by zengargoyle at 3:16 PM on December 7, 2019 [1 favorite]
I learned completing the square in Algebra II in 10th grade. I couldn't do it now to save my life. I do remember the quadratic equation being beat into my head by Mr. Bernath that year. Almost 30 years later, I remember it. That might be the only thing I remember from high school math.
posted by kathrynm at 7:37 PM on December 7, 2019
posted by kathrynm at 7:37 PM on December 7, 2019
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posted by otherchaz at 8:18 PM on December 6, 2019